How do you integrate #int 1/sqrt(9x^2-18x-7) # using trigonometric substitution?

1 Answer
Liam P.
May 4, 2018

#-1/12ln|csctheta+cottheta|+C#

Explanation:

First, you can complete the square in the bottom

#int1/sqrt((3x-3)^2-16)dx#

Then, use the substitution #3x-3=4sectheta#
Differentiating this gives #3dx=4secthetatantheta*d(theta)#
Substituting this in gives:
#int1/sqrt(16sec^2theta-16)*4/3secthetatantheta*d(theta)#
Since #1+tan^2theta=sec^2theta#
#=1/3intsectheta*d(theta)#
#=1/3ln|csctheta+cottheta|+C#
Then, you can draw a right triangle based on your intial substitution to turn #theta# back into #x#.

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